Question 9.17: Calculate the value of output voltage in terms of the input ...

Applying the superposition theorem, we calculate the output voltage for each of the input voltages and
then determine the combined effect.
(i) When V_{1} is applied and V_{2} = V_{3} = 0.
V_{o1}= – \frac{4R}{R} V_{1}= – 4V_{1}
(ii) When V_{2} is applied and V_{1} = V_{3} = 0
Considering the two 3R resistances in parallel (as V_{3} is also at zero potential, i.e. earthed), the
equivalent resistance is 1.5 R.
V_{A}=\frac{V_{2} ×1.5 R}{2R+10.5R}=\frac{V_{2} × 1.5 R}{3.5 R}=\frac{3}{7}  V_{2}.
Let the output voltage is  V_{o2}.
V_{o2}=\left\lgroup 1+\frac{R_{f}}{R}\right\rgroup V_{A}=\left\lgroup1+\frac{4R}{R}\right\rgroup V_{A}
=5×\frac{3}{7}  V_{2} =\frac{15}{7}  V_{2}
(iii) V_{3} is applied, V_{1} = V_{2} = 0 (at ground potential. The connections are show)
=\frac{6/5R}{21/5R}  V_{3} =\frac{2}{7}  V_{3}
V_{o3}=\left\lgroup 1+ \frac{4R}{R} \right\rgroup VA
=5×\frac{2}{7}  V_{3} =\frac{10}{7}  V_{3}
The total output              =V_{o1}+V_{o2}+V_{o3}
V_{o}=\left[-4V_{1}+\frac{15}{7}V_{2}+\frac{2}{7}V_{3}\right]